Question

if x+y+z=1 , xy+yx+zx = -1 and xyz=-1 then find the value of x^3+y^3+z^3

Answer 1

any1 help me

Answer 2

\[x+y+z = 1\]

\[xy+yx+zx = -1\]

\[xyz = -1\]

this is right?

Answer 3

yea tanvi that is right

Answer 4

in the second one, xy+yx = 2xy

Answer 5

you want to get (x+y+z)^3 and write it out:\[(x+y+z)^3=x^3+y^3+z^3+3(x^2y+x^2z+y^2z+xy^2+xz^2+yz^2)+6xyz\]

Answer 6

find the value of \[x^3+y^3+z^3\]

Answer 7

yes joseph sir what after then ?

Answer 8

that means one of x, y or z is a one. the others can be any twp numbers that are reciprocals of each other.

Answer 9

then playing around with the terms we get:
\[x^3+y^3+z^3+3(x^2y+x^2z+y^2z+xy^2+xz^2+yz^2+3xyz)-3xyz\]
\[x^3+y^3+z^3+3(x+y+z)(xy+yz+xz)-3xyz\]

Answer 10

oh yes and then easily we can put the values i got it joesh sir thanks very much

Answer 11

sir r u on facebook ?

Answer 12

so we now have:\[(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(xy+yz+xz)-3xyz\]
just put in the values and solve for your sum.

Answer 13

Professor Joe

Answer 14

for completion sake:
\[(1)^3 = x^3+y^3+z^3+3(1)(-1)-3(-1) \iff x^3+y^3+z^3=1+3-3 = 1\]

Answer 15

joseph sir r u on facebook ?

Answer 16

Im not sure =/ i had an account a looooong time ago, but when i started working at a high school i stopped using it (job rules)

Answer 17

oh yes u r right sir ...........

Answer 18

btw, if you wanted the actual solutions to this system, its the same as solving the cubic equation:
\[x^3-x^2-x+1=0\]The roots to this equations are the solutions to this system.

Answer 19

joseph sir help me here : http://openstudy.com/groups/mathematics/updates/4e5f63e70b8b1f45b49de337

Answer 20

\[x^2(x-1)-(x-1)=0\iff (x^2-1)(x-1)=0\iff (x+1)(x-1)^2=0\]
so x=-1, y=1, z =1 (and any permutation) is a solution to the system

Answer 21

thanks

Answer 22

joe can you help me out

Answer 23

i can try, whats the question?

Answer 24

http://openstudy.com/groups/mathematics/updates/4e5f61d80b8b1f45b49de18e#/groups/mathematics/updates/4e5f61d80b8b1f45b49de18e